Metamath Proof Explorer


Theorem mirln

Description: If two points are on the same line, so is the mirror point of one through the other. (Contributed by Thierry Arnoux, 21-Dec-2019)

Ref Expression
Hypotheses mirval.p ⊢ P = Base G
mirval.d ⊢ - ˙ = dist ⁡ G
mirval.i ⊢ I = Itv ⁡ G
mirval.l ⊢ L = Line 𝒢 ⁡ G
mirval.s ⊢ S = pInv 𝒢 ⁡ G
mirval.g ⊢ φ → G ∈ 𝒢 Tarski
mirln.m ⊢ M = S ⁡ A
mirln.1 ⊢ φ → D ∈ ran ⁡ L
mirln.a ⊢ φ → A ∈ D
mirln.b ⊢ φ → B ∈ D
Assertion mirln ⊢ φ → M ⁡ B ∈ D

Proof

Step Hyp Ref Expression
1 mirval.p ⊢ P = Base G
2 mirval.d ⊢ - ˙ = dist ⁡ G
3 mirval.i ⊢ I = Itv ⁡ G
4 mirval.l ⊢ L = Line 𝒢 ⁡ G
5 mirval.s ⊢ S = pInv 𝒢 ⁡ G
6 mirval.g ⊢ φ → G ∈ 𝒢 Tarski
7 mirln.m ⊢ M = S ⁡ A
8 mirln.1 ⊢ φ → D ∈ ran ⁡ L
9 mirln.a ⊢ φ → A ∈ D
10 mirln.b ⊢ φ → B ∈ D
11 simpr ⊢ φ ∧ A = B → A = B
12 11 fveq2d ⊢ φ ∧ A = B → M ⁡ A = M ⁡ B
13 6 adantr ⊢ φ ∧ A = B → G ∈ 𝒢 Tarski
14 1 4 3 6 8 9 tglnpt ⊢ φ → A ∈ P
15 14 adantr ⊢ φ ∧ A = B → A ∈ P
16 1 2 3 4 5 13 15 7 mircinv ⊢ φ ∧ A = B → M ⁡ A = A
17 12 16 eqtr3d ⊢ φ ∧ A = B → M ⁡ B = A
18 9 adantr ⊢ φ ∧ A = B → A ∈ D
19 17 18 eqeltrd ⊢ φ ∧ A = B → M ⁡ B ∈ D
20 6 adantr ⊢ φ ∧ A ≠ B → G ∈ 𝒢 Tarski
21 14 adantr ⊢ φ ∧ A ≠ B → A ∈ P
22 1 4 3 6 8 10 tglnpt ⊢ φ → B ∈ P
23 22 adantr ⊢ φ ∧ A ≠ B → B ∈ P
24 1 2 3 4 5 20 21 7 23 mircl ⊢ φ ∧ A ≠ B → M ⁡ B ∈ P
25 simpr ⊢ φ ∧ A ≠ B → A ≠ B
26 1 2 3 4 5 6 14 7 22 mirbtwn ⊢ φ → A ∈ M ⁡ B I B
27 26 adantr ⊢ φ ∧ A ≠ B → A ∈ M ⁡ B I B
28 1 3 4 20 21 23 24 25 27 btwnlng2 ⊢ φ ∧ A ≠ B → M ⁡ B ∈ A L B
29 8 adantr ⊢ φ ∧ A ≠ B → D ∈ ran ⁡ L
30 9 adantr ⊢ φ ∧ A ≠ B → A ∈ D
31 10 adantr ⊢ φ ∧ A ≠ B → B ∈ D
32 1 3 4 20 21 23 25 25 29 30 31 tglinethru ⊢ φ ∧ A ≠ B → D = A L B
33 28 32 eleqtrrd ⊢ φ ∧ A ≠ B → M ⁡ B ∈ D
34 19 33 pm2.61dane ⊢ φ → M ⁡ B ∈ D